3.92.65 \(\int \frac {-8+(-x-2 x^2) \log (\frac {1}{2} (4 x^2-x^2 \log (3))) \log ^2(\log (\frac {1}{2} (4 x^2-x^2 \log (3))))}{16 x \log (\frac {1}{2} (4 x^2-x^2 \log (3)))+(-8 x-8 x^2-8 x^3) \log (\frac {1}{2} (4 x^2-x^2 \log (3))) \log (\log (\frac {1}{2} (4 x^2-x^2 \log (3))))+(x+2 x^2+3 x^3+2 x^4+x^5) \log (\frac {1}{2} (4 x^2-x^2 \log (3))) \log ^2(\log (\frac {1}{2} (4 x^2-x^2 \log (3))))} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{1+x (1+x)-\frac {4}{\log \left (\log \left (\frac {1}{2} x^2 (4-\log (3))\right )\right )}} \]

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Rubi [F]  time = 51.56, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-8+\left (-x-2 x^2\right ) \log \left (\frac {1}{2} \left (4 x^2-x^2 \log (3)\right )\right ) \log ^2\left (\log \left (\frac {1}{2} \left (4 x^2-x^2 \log (3)\right )\right )\right )}{16 x \log \left (\frac {1}{2} \left (4 x^2-x^2 \log (3)\right )\right )+\left (-8 x-8 x^2-8 x^3\right ) \log \left (\frac {1}{2} \left (4 x^2-x^2 \log (3)\right )\right ) \log \left (\log \left (\frac {1}{2} \left (4 x^2-x^2 \log (3)\right )\right )\right )+\left (x+2 x^2+3 x^3+2 x^4+x^5\right ) \log \left (\frac {1}{2} \left (4 x^2-x^2 \log (3)\right )\right ) \log ^2\left (\log \left (\frac {1}{2} \left (4 x^2-x^2 \log (3)\right )\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8 + (-x - 2*x^2)*Log[(4*x^2 - x^2*Log[3])/2]*Log[Log[(4*x^2 - x^2*Log[3])/2]]^2)/(16*x*Log[(4*x^2 - x^2*
Log[3])/2] + (-8*x - 8*x^2 - 8*x^3)*Log[(4*x^2 - x^2*Log[3])/2]*Log[Log[(4*x^2 - x^2*Log[3])/2]] + (x + 2*x^2
+ 3*x^3 + 2*x^4 + x^5)*Log[(4*x^2 - x^2*Log[3])/2]*Log[Log[(4*x^2 - x^2*Log[3])/2]]^2),x]

[Out]

(1 + x + x^2)^(-1) - 16*Defer[Int][1/((1 + x + x^2)^2*(4 - (1 + x + x^2)*Log[Log[-1/2*(x^2*(-4 + Log[3]))]])^2
), x] - 32*Defer[Int][x/((1 + x + x^2)^2*(4 - (1 + x + x^2)*Log[Log[-1/2*(x^2*(-4 + Log[3]))]])^2), x] - 8*Def
er[Int][1/(x*Log[-1/2*(x^2*(-4 + Log[3]))]*(4 - (1 + x + x^2)*Log[Log[-1/2*(x^2*(-4 + Log[3]))]])^2), x] - 8*D
efer[Int][1/((1 + x + x^2)^2*(-4 + (1 + x + x^2)*Log[Log[-1/2*(x^2*(-4 + Log[3]))]])), x] - 16*Defer[Int][x/((
1 + x + x^2)^2*(-4 + (1 + x + x^2)*Log[Log[-1/2*(x^2*(-4 + Log[3]))]])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-8-x (1+2 x) \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right ) \log ^2\left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )}{x \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right ) \left (4-\left (1+x+x^2\right ) \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )^2} \, dx\\ &=\int \left (\frac {-1-2 x}{\left (1+x+x^2\right )^2}-\frac {8 \left (1+2 x+3 x^2+2 x^3+x^4+2 x \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )+4 x^2 \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )}{x \left (1+x+x^2\right )^2 \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right ) \left (-4+\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x^2 \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )^2}-\frac {8 (1+2 x)}{\left (1+x+x^2\right )^2 \left (-4+\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x^2 \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )}\right ) \, dx\\ &=-\left (8 \int \frac {1+2 x+3 x^2+2 x^3+x^4+2 x \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )+4 x^2 \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )}{x \left (1+x+x^2\right )^2 \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right ) \left (-4+\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x^2 \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )^2} \, dx\right )-8 \int \frac {1+2 x}{\left (1+x+x^2\right )^2 \left (-4+\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x^2 \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )} \, dx+\int \frac {-1-2 x}{\left (1+x+x^2\right )^2} \, dx\\ &=\frac {1}{1+x+x^2}-8 \int \frac {\left (1+x+x^2\right )^2+2 x (1+2 x) \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )}{x \left (1+x+x^2\right )^2 \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right ) \left (4-\left (1+x+x^2\right ) \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )^2} \, dx-8 \int \frac {-1-2 x}{\left (1+x+x^2\right )^2 \left (4-\left (1+x+x^2\right ) \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )} \, dx\\ &=\frac {1}{1+x+x^2}-8 \int \left (\frac {1+2 x+3 x^2+2 x^3+x^4+2 x \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )+4 x^2 \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )}{x \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right ) \left (-4+\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x^2 \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )^2}-\frac {(1+x) \left (1+2 x+3 x^2+2 x^3+x^4+2 x \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )+4 x^2 \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )}{\left (1+x+x^2\right )^2 \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right ) \left (-4+\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x^2 \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )^2}-\frac {(1+x) \left (1+2 x+3 x^2+2 x^3+x^4+2 x \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )+4 x^2 \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )}{\left (1+x+x^2\right ) \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right ) \left (-4+\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x^2 \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )^2}\right ) \, dx-8 \int \left (\frac {1}{\left (1+x+x^2\right )^2 \left (-4+\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x^2 \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )}+\frac {2 x}{\left (1+x+x^2\right )^2 \left (-4+\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x^2 \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )}\right ) \, dx\\ &=\frac {1}{1+x+x^2}-8 \int \frac {1+2 x+3 x^2+2 x^3+x^4+2 x \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )+4 x^2 \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )}{x \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right ) \left (-4+\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x^2 \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )^2} \, dx+8 \int \frac {(1+x) \left (1+2 x+3 x^2+2 x^3+x^4+2 x \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )+4 x^2 \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )}{\left (1+x+x^2\right )^2 \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right ) \left (-4+\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x^2 \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )^2} \, dx+8 \int \frac {(1+x) \left (1+2 x+3 x^2+2 x^3+x^4+2 x \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )+4 x^2 \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )}{\left (1+x+x^2\right ) \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right ) \left (-4+\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x^2 \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )^2} \, dx-8 \int \frac {1}{\left (1+x+x^2\right )^2 \left (-4+\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x^2 \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )} \, dx-16 \int \frac {x}{\left (1+x+x^2\right )^2 \left (-4+\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )+x^2 \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )} \, dx\\ &=\frac {1}{1+x+x^2}-8 \int \frac {\left (1+x+x^2\right )^2+2 x (1+2 x) \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )}{x \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right ) \left (4-\left (1+x+x^2\right ) \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )^2} \, dx+8 \int \frac {(1+x) \left (\left (1+x+x^2\right )^2+2 x (1+2 x) \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )}{\left (1+x+x^2\right )^2 \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right ) \left (4-\left (1+x+x^2\right ) \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )^2} \, dx+8 \int \frac {(1+x) \left (\left (1+x+x^2\right )^2+2 x (1+2 x) \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )}{\left (1+x+x^2\right ) \log \left (-\frac {1}{2} x^2 (-4+\log (3))\right ) \left (4-\left (1+x+x^2\right ) \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )^2} \, dx-8 \int \frac {1}{\left (1+x+x^2\right )^2 \left (-4+\left (1+x+x^2\right ) \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )} \, dx-16 \int \frac {x}{\left (1+x+x^2\right )^2 \left (-4+\left (1+x+x^2\right ) \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.67, size = 38, normalized size = 1.36 \begin {gather*} \frac {\log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )}{-4+\left (1+x+x^2\right ) \log \left (\log \left (-\frac {1}{2} x^2 (-4+\log (3))\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 + (-x - 2*x^2)*Log[(4*x^2 - x^2*Log[3])/2]*Log[Log[(4*x^2 - x^2*Log[3])/2]]^2)/(16*x*Log[(4*x^2
- x^2*Log[3])/2] + (-8*x - 8*x^2 - 8*x^3)*Log[(4*x^2 - x^2*Log[3])/2]*Log[Log[(4*x^2 - x^2*Log[3])/2]] + (x +
2*x^2 + 3*x^3 + 2*x^4 + x^5)*Log[(4*x^2 - x^2*Log[3])/2]*Log[Log[(4*x^2 - x^2*Log[3])/2]]^2),x]

[Out]

Log[Log[-1/2*(x^2*(-4 + Log[3]))]]/(-4 + (1 + x + x^2)*Log[Log[-1/2*(x^2*(-4 + Log[3]))]])

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fricas [A]  time = 0.60, size = 42, normalized size = 1.50 \begin {gather*} \frac {\log \left (\log \left (-\frac {1}{2} \, x^{2} \log \relax (3) + 2 \, x^{2}\right )\right )}{{\left (x^{2} + x + 1\right )} \log \left (\log \left (-\frac {1}{2} \, x^{2} \log \relax (3) + 2 \, x^{2}\right )\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-x)*log(-1/2*x^2*log(3)+2*x^2)*log(log(-1/2*x^2*log(3)+2*x^2))^2-8)/((x^5+2*x^4+3*x^3+2*x^2+
x)*log(-1/2*x^2*log(3)+2*x^2)*log(log(-1/2*x^2*log(3)+2*x^2))^2+(-8*x^3-8*x^2-8*x)*log(-1/2*x^2*log(3)+2*x^2)*
log(log(-1/2*x^2*log(3)+2*x^2))+16*x*log(-1/2*x^2*log(3)+2*x^2)),x, algorithm="fricas")

[Out]

log(log(-1/2*x^2*log(3) + 2*x^2))/((x^2 + x + 1)*log(log(-1/2*x^2*log(3) + 2*x^2)) - 4)

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giac [B]  time = 1.57, size = 140, normalized size = 5.00 \begin {gather*} \frac {4}{x^{4} \log \left (-\log \relax (2) + \log \left (-x^{2} \log \relax (3) + 4 \, x^{2}\right )\right ) + 2 \, x^{3} \log \left (-\log \relax (2) + \log \left (-x^{2} \log \relax (3) + 4 \, x^{2}\right )\right ) + 3 \, x^{2} \log \left (-\log \relax (2) + \log \left (-x^{2} \log \relax (3) + 4 \, x^{2}\right )\right ) - 4 \, x^{2} + 2 \, x \log \left (-\log \relax (2) + \log \left (-x^{2} \log \relax (3) + 4 \, x^{2}\right )\right ) - 4 \, x + \log \left (-\log \relax (2) + \log \left (-x^{2} \log \relax (3) + 4 \, x^{2}\right )\right ) - 4} + \frac {1}{x^{2} + x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-x)*log(-1/2*x^2*log(3)+2*x^2)*log(log(-1/2*x^2*log(3)+2*x^2))^2-8)/((x^5+2*x^4+3*x^3+2*x^2+
x)*log(-1/2*x^2*log(3)+2*x^2)*log(log(-1/2*x^2*log(3)+2*x^2))^2+(-8*x^3-8*x^2-8*x)*log(-1/2*x^2*log(3)+2*x^2)*
log(log(-1/2*x^2*log(3)+2*x^2))+16*x*log(-1/2*x^2*log(3)+2*x^2)),x, algorithm="giac")

[Out]

4/(x^4*log(-log(2) + log(-x^2*log(3) + 4*x^2)) + 2*x^3*log(-log(2) + log(-x^2*log(3) + 4*x^2)) + 3*x^2*log(-lo
g(2) + log(-x^2*log(3) + 4*x^2)) - 4*x^2 + 2*x*log(-log(2) + log(-x^2*log(3) + 4*x^2)) - 4*x + log(-log(2) + l
og(-x^2*log(3) + 4*x^2)) - 4) + 1/(x^2 + x + 1)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (-2 x^{2}-x \right ) \ln \left (-\frac {x^{2} \ln \relax (3)}{2}+2 x^{2}\right ) \ln \left (\ln \left (-\frac {x^{2} \ln \relax (3)}{2}+2 x^{2}\right )\right )^{2}-8}{\left (x^{5}+2 x^{4}+3 x^{3}+2 x^{2}+x \right ) \ln \left (-\frac {x^{2} \ln \relax (3)}{2}+2 x^{2}\right ) \ln \left (\ln \left (-\frac {x^{2} \ln \relax (3)}{2}+2 x^{2}\right )\right )^{2}+\left (-8 x^{3}-8 x^{2}-8 x \right ) \ln \left (-\frac {x^{2} \ln \relax (3)}{2}+2 x^{2}\right ) \ln \left (\ln \left (-\frac {x^{2} \ln \relax (3)}{2}+2 x^{2}\right )\right )+16 x \ln \left (-\frac {x^{2} \ln \relax (3)}{2}+2 x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2-x)*ln(-1/2*x^2*ln(3)+2*x^2)*ln(ln(-1/2*x^2*ln(3)+2*x^2))^2-8)/((x^5+2*x^4+3*x^3+2*x^2+x)*ln(-1/2*
x^2*ln(3)+2*x^2)*ln(ln(-1/2*x^2*ln(3)+2*x^2))^2+(-8*x^3-8*x^2-8*x)*ln(-1/2*x^2*ln(3)+2*x^2)*ln(ln(-1/2*x^2*ln(
3)+2*x^2))+16*x*ln(-1/2*x^2*ln(3)+2*x^2)),x)

[Out]

int(((-2*x^2-x)*ln(-1/2*x^2*ln(3)+2*x^2)*ln(ln(-1/2*x^2*ln(3)+2*x^2))^2-8)/((x^5+2*x^4+3*x^3+2*x^2+x)*ln(-1/2*
x^2*ln(3)+2*x^2)*ln(ln(-1/2*x^2*ln(3)+2*x^2))^2+(-8*x^3-8*x^2-8*x)*ln(-1/2*x^2*ln(3)+2*x^2)*ln(ln(-1/2*x^2*ln(
3)+2*x^2))+16*x*ln(-1/2*x^2*ln(3)+2*x^2)),x)

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maxima [A]  time = 0.52, size = 46, normalized size = 1.64 \begin {gather*} \frac {\log \left (-\log \relax (2) + 2 \, \log \relax (x) + \log \left (-\log \relax (3) + 4\right )\right )}{{\left (x^{2} + x + 1\right )} \log \left (-\log \relax (2) + 2 \, \log \relax (x) + \log \left (-\log \relax (3) + 4\right )\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-x)*log(-1/2*x^2*log(3)+2*x^2)*log(log(-1/2*x^2*log(3)+2*x^2))^2-8)/((x^5+2*x^4+3*x^3+2*x^2+
x)*log(-1/2*x^2*log(3)+2*x^2)*log(log(-1/2*x^2*log(3)+2*x^2))^2+(-8*x^3-8*x^2-8*x)*log(-1/2*x^2*log(3)+2*x^2)*
log(log(-1/2*x^2*log(3)+2*x^2))+16*x*log(-1/2*x^2*log(3)+2*x^2)),x, algorithm="maxima")

[Out]

log(-log(2) + 2*log(x) + log(-log(3) + 4))/((x^2 + x + 1)*log(-log(2) + 2*log(x) + log(-log(3) + 4)) - 4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {\ln \left (2\,x^2-\frac {x^2\,\ln \relax (3)}{2}\right )\,\left (2\,x^2+x\right )\,{\ln \left (\ln \left (2\,x^2-\frac {x^2\,\ln \relax (3)}{2}\right )\right )}^2+8}{\ln \left (2\,x^2-\frac {x^2\,\ln \relax (3)}{2}\right )\,\left (x^5+2\,x^4+3\,x^3+2\,x^2+x\right )\,{\ln \left (\ln \left (2\,x^2-\frac {x^2\,\ln \relax (3)}{2}\right )\right )}^2-\ln \left (2\,x^2-\frac {x^2\,\ln \relax (3)}{2}\right )\,\left (8\,x^3+8\,x^2+8\,x\right )\,\ln \left (\ln \left (2\,x^2-\frac {x^2\,\ln \relax (3)}{2}\right )\right )+16\,x\,\ln \left (2\,x^2-\frac {x^2\,\ln \relax (3)}{2}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(2*x^2 - (x^2*log(3))/2))^2*log(2*x^2 - (x^2*log(3))/2)*(x + 2*x^2) + 8)/(16*x*log(2*x^2 - (x^2*l
og(3))/2) - log(log(2*x^2 - (x^2*log(3))/2))*log(2*x^2 - (x^2*log(3))/2)*(8*x + 8*x^2 + 8*x^3) + log(log(2*x^2
- (x^2*log(3))/2))^2*log(2*x^2 - (x^2*log(3))/2)*(x + 2*x^2 + 3*x^3 + 2*x^4 + x^5)),x)

[Out]

int(-(log(log(2*x^2 - (x^2*log(3))/2))^2*log(2*x^2 - (x^2*log(3))/2)*(x + 2*x^2) + 8)/(16*x*log(2*x^2 - (x^2*l
og(3))/2) - log(log(2*x^2 - (x^2*log(3))/2))*log(2*x^2 - (x^2*log(3))/2)*(8*x + 8*x^2 + 8*x^3) + log(log(2*x^2
- (x^2*log(3))/2))^2*log(2*x^2 - (x^2*log(3))/2)*(x + 2*x^2 + 3*x^3 + 2*x^4 + x^5)), x)

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sympy [B]  time = 0.40, size = 54, normalized size = 1.93 \begin {gather*} \frac {4}{- 4 x^{2} - 4 x + \left (x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 1\right ) \log {\left (\log {\left (- \frac {x^{2} \log {\relax (3 )}}{2} + 2 x^{2} \right )} \right )} - 4} + \frac {1}{x^{2} + x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2-x)*ln(-1/2*x**2*ln(3)+2*x**2)*ln(ln(-1/2*x**2*ln(3)+2*x**2))**2-8)/((x**5+2*x**4+3*x**3+2*
x**2+x)*ln(-1/2*x**2*ln(3)+2*x**2)*ln(ln(-1/2*x**2*ln(3)+2*x**2))**2+(-8*x**3-8*x**2-8*x)*ln(-1/2*x**2*ln(3)+2
*x**2)*ln(ln(-1/2*x**2*ln(3)+2*x**2))+16*x*ln(-1/2*x**2*ln(3)+2*x**2)),x)

[Out]

4/(-4*x**2 - 4*x + (x**4 + 2*x**3 + 3*x**2 + 2*x + 1)*log(log(-x**2*log(3)/2 + 2*x**2)) - 4) + 1/(x**2 + x + 1
)

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